Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $r = \dfrac{6q + 42}{-10q - 40} \times \dfrac{q + 4}{q^2 + 8q + 7} $
Answer: First factor the quadratic. $r = \dfrac{6q + 42}{-10q - 40} \times \dfrac{q + 4}{(q + 7)(q + 1)} $ Then factor out any other terms. $r = \dfrac{6(q + 7)}{-10(q + 4)} \times \dfrac{q + 4}{(q + 7)(q + 1)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 6(q + 7) \times (q + 4) } { -10(q + 4) \times (q + 7)(q + 1) } $ $r = \dfrac{ 6(q + 7)(q + 4)}{ -10(q + 4)(q + 7)(q + 1)} $ Notice that $(q + 4)$ and $(q + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 6\cancel{(q + 7)}(q + 4)}{ -10(q + 4)\cancel{(q + 7)}(q + 1)} $ We are dividing by $q + 7$ , so $q + 7 \neq 0$ Therefore, $q \neq -7$ $r = \dfrac{ 6\cancel{(q + 7)}\cancel{(q + 4)}}{ -10\cancel{(q + 4)}\cancel{(q + 7)}(q + 1)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $r = \dfrac{6}{-10(q + 1)} $ $r = \dfrac{-3}{5(q + 1)} ; \space q \neq -7 ; \space q \neq -4 $